Note:-this is the case when we r ignoring the mod.....
we can clearly see,as mod is applied in the question,,,,so,
negative sign will become positive and we can say
|-x|+|-sinx|=x+sinx
so
LHL =RHL
Hence limit will be continuous at
now if we differentiate the limit at x =0
then we get
LHL=-1-cos0=-1-1=-2
RHL=1+cos0=2
Clearly we can see that
LHL is not eqaul to RHL
so the limit will not differentiable at x=0
{hope it helps}
......................
.......
also don't check LHL and RHL of dy/dx of
to make sure that function is differentiable at every point.That isn't always true
ALWAYS CHECK BY FIRST PRINCIPLE
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Answers & Comments
Note:-this is the case when we r ignoring the mod.....
we can clearly see,as mod is applied in the question,,,,so,
negative sign will become positive and we can say
|-x|+|-sinx|=x+sinx
so
LHL =RHL
Hence limit will be continuous at
now if we differentiate the limit at x =0
then we get
LHL=-1-cos0=-1-1=-2
RHL=1+cos0=2
Clearly we can see that
LHL is not eqaul to RHL
so the limit will not differentiable at x=0
{hope it helps}
......................
.......
also don't check LHL and RHL of dy/dx of
to make sure that function is differentiable at every point.That isn't always true
ALWAYS CHECK BY FIRST PRINCIPLE